<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Using the result \(\sum_{r=2}^{n} \frac{2}{(2r + 1)(2r + 3)} = \frac{1}{5} - \frac{1}{2n + 3}\), find \(\sum_{r=2}^{2n} \frac{2}{(2r - 1)(2r + 1)}\) in terms of \(n\).</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">A sequence of positive numbers \(x_1, x_2, x_3, \dots\) satisfies the recurrence relation\[x_{n+1} = \sqrt{4 + 5x_n} \quad \text{for } n = 1, 2, 3, \dots\]</div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Given that the sequence converges to \(L\), find the exact value of \(L\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Prove that&nbsp;\(x_{n+1}^2 - L^2 = 5(x_n - L)\).</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Use the result in part (ii) to show that if&nbsp;\(x_n &gt; L\), then&nbsp;\(x_{n+1} &gt; L\).</span><span class="inline-marks">[2]</span></div></li></ol></li></ol>

Using the result \(\sum_{r=2}^{n} \frac{2}{(2r + 1)(2r + 3)} = \frac{1}{5} - \frac{1}{2n + 3}\), find \(\sum_{r=2}^{2n} \frac{2}{(2r - 1)(2r + 1)}\) in terms of \(n\).

A sequence of positive numbers \(x_1, x_2, x_3, \dots\) satisfies the recurrence relation\n\[x_{n+1} = \sqrt{4 + 5x_n} \quad \text{for } n = 1, 2, 3, \dots\]

A sequence of positive numbers \(x_1, x_2, x_3, \dots\) satisfies the recurrence relation\[x_{n+1} = \sqrt{4 + 5x_n} \quad \text{for } n = 1, 2, 3, \dots\]

Given that the sequence converges to \(L\), find the exact value of \(L\).

Prove that \(x_{n+1}^2 - L^2 = 5(x_n - L)\).

Use the result in part (ii) to show that if \(x_n > L\), then \(x_{n+1} > L\).

Recurrence Relations and Summation

H2 Math

International Baccalaureate Higher Level Mathematics

2025 TJC P2 Q2

2025 TJC P2 Q2